The two linear functions are shown below f(x)=5/6x+3 what is true ?

Mathematics

1 answer:

Deffense [45]3 years ago

8 0

What are the two functions?

You might be interested in

More Calculus! (I'm so sorry)

Olenka [21]

Recall that converting from Cartesian to polar coordinates involves the identities

$\begin{cases}y(r,\phi)=r\sin\phi\\x(r,\phi)=r\cos\phi\end{cases}$

As a function in polar coordinates, $r$ depends on $\phi$ , so you can write $r=r(\phi)$ .

Differentiating the identities with respect to $\phi$ gives

$\begin{cases}\dfrac{\mathrm dy}{\mathrm d\phi}=\dfrac{\mathrm dr}{\mathrm d\phi}\sin\phi+r\cos\phi\\\\\dfrac{\mathrm dx}{\mathrm d\phi}=\dfrac{\mathrm dr}{\mathrm d\phi}\cos\phi-r\sin\phi\end{cases}$

The slope of the tangent line to $r(\phi)$ is given by

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\phi}}{\frac{\mathrm dx}{\mathrm d\phi}}=\dfrac{\frac{\mathrm dr}{\mathrm d\phi}\sin\phi+r\cos\phi}{\frac{\mathrm dr}{\mathrm d\phi}\cos\phi-r\sin\phi}$

Given $r(\phi)=3\cos\phi$ , you have $\dfrac{\mathrm dr}{\mathrm d\phi}=-3\sin\phi$ . So the tangent line to $r(\phi)$ has a slope of

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-3\sin^2\phi+3\cos^2\phi}{-3\sin\phi\cos\phi-3\cos\phi\sin\phi}=\dfrac{3\cos2\phi}{-3\sin2\phi}=-\cot2\phi$

When $\phi=120^\circ=\dfrac{2\pi}3\text{ rad}$ , the tangent line has slope

$\dfrac{\mathrm dy}{\mathrm dx}=-\cot\dfrac{4\pi}3=-\dfrac1{\sqrt3}$

This line is tangent to the point $(r,\phi)=\left(-\dfrac32,\dfrac{2\pi}3\right)$ which in Cartesian coordinates is equivalent to $(x,y)=\left(\dfrac34,-\dfrac{3\sqrt3}4\right)$ , so the equation of the tangent line is

$y+\dfrac{3\sqrt3}4=-\dfrac1{\sqrt3}\left(x-\dfrac34\right)$

In polar coordinates, this line has equation

$r\sin\phi+\dfrac{3\sqrt3}4=-\dfrac1{\sqrt3}\left(r\cos\phi-\dfrac34\right)$
$\implies r=-\dfrac{3\sqrt3}{2\sqrt3\cos\phi+6\sin\phi}$

The tangent line passes through the y-axis when $x=0$ , so the y-intercept is $\left(0,-\dfrac{\sqrt3}2\right)$ .

The vector from this point to the point of tangency on $r(\phi)$ is given by the difference of the vector from the origin to the y-intercept (which I'll denote $\mathbf a$ ) and the vector from the origin to the point of tangency (denoted by $\mathbf b$ ). In the attached graphic, this corresponds to the green arrow.

$\mathbf b-\mathbf a=\left(\dfrac34,-\dfrac{3\sqrt3}4\right)-\left(0,-\dfrac{\sqrt3}2\right)=\left(\dfrac34,-\dfrac{\sqrt3}4\right)$

The angle between this vector and the vector pointing to the point of tangency is what you're looking for. This is given by

$\mathbf b\cdot(\mathbf b-\mathbf a)=\|\mathbf b\|\|\mathbf b-\mathbf a\|\cos\theta$
$\dfrac98=\dfrac{3\sqrt3}4\cos\theta$
$\implies\theta=\dfrac\pi6\text{ rad}=30^\circ$

The second problem is just a matter of computing the second derivative of $\phi$ with respect to $t$ and plugging in $t=2$ .

$\phi(t)=2t^3-6t$
$\phi'(t)=6t^2-6$
$\phi''(t)=12t$
$\implies\phi''(2)=24$